Lattices in $\mathbb C$ as modules of the ring of integers in an imaginary quadratic field

Question

Let $K$ be an imaginary quadratic number field and let $O_K\subset K$ be the ring of algebraic integers in $K$. Let us call a lattice $\Lambda\subset\mathbb C$ normalized if the tori $\mathbb C/\Lambda$ and $\mathbb C/O_K$ have the same area.

The following is true:

A normalized lattice $\Lambda\subset\mathbb C$ is a $O_K$-module (i.e. $\alpha\lambda\in\Lambda$ for all $\alpha\in O_K$, $\lambda\in\Lambda$) if and only if $|\lambda|^2\in\mathbb Z$ for every $\lambda\in\Lambda$.

What is a proof of this statement (if possible direct/elegant/short)? Are there any similar/analogous results?

It is easy to see the $\Rightarrow$ implication: if a normalized lattice $\Lambda\subset\mathbb C$ is a $O_K$ module then for every $\lambda\in\Lambda$ we see that $\lambda O_K\subset\Lambda$, and since $\Lambda$ is normalized, we get (if $\lambda\neq0$) $|\lambda|^2=[\Lambda:\lambda O_K]$, hence $|\lambda|^2\in\mathbb Z$. What is, however, a proof of the $\Leftarrow$ implication? (I can prove it with a random-looking calculation, but a conceptual proof should exist.)

context: The statement is equivalent to the description of the ideal class group of $K$ in terms of quadratic forms (any $\Lambda$ gives us a quadratic form, namely $\lambda\mapsto|\lambda|^2$). I just want to understand this description and its proof geometrically (so everything should be well-known, just not to me).

Answer 1

I am afraid that I do not know a "direct/elegant/short" proof of the implication that you are asking about. For me, the "real reason" that it is true, is not geometric but algebraic, and maybe more sophisticated than it should be.

Your $\Lambda$ carries an integral-valued quadratic form, and I would start by proving that$$\mathbb{Q} \Lambda := \{\text{rational multiples of }\Lambda\}$$is a $1$-dimensional $K$-vector space inside the field of complex numbers; this is done by considering the determinant of the inner product on $\Lambda$. Next I would consider the ring $R$ (say) of all $x$ in $K$ with $x\Lambda \subset \Lambda$. That is a subring of your $\mathcal{O}_K$ of finite index. Now there is a subtlety, which derives from an algebraic property of $R$ (it is a Gorenstein ring) but that can also be proved by an unpleasant explicit calculation (I saw it once in Borevich and Shafarevich's book on number theory), which says that $\Lambda$ is "locally" generated by one element over $R$. In explicit terms, this means that for every positive integer $n$ there exists $\lambda$ in $\Lambda$ for which $[\Lambda : R\lambda]$ is coprime to $n$. Now, by your own calculation,$$|\lambda|^2 = [\Lambda : R\lambda]\, {{\text{covol}(\Lambda)}\over{\text{covol}(R)}} = {{[\Lambda : R\lambda]}\over{[\mathcal{O}_K: R]}}.$$Choosing $n = [\mathcal{O}_K : R]$ and using that $|\lambda|^2$ is an integer, one deduces that $\mathcal{O}_K = R$, as required.

The Gorenstein property derives from $R$ being monogenic over the ring $\mathbb{Z}$ of integers. It is a specific property for quadratic rings, and makes it hard to generalize the proof and probably even the statement to higher dimensional situations.